Further results on the distinctness of modulo 2 reductions of primitive sequences over $\mathbf{Z}/(2^{32}-1)$

Abstract

Recently, primitive sequences over $\mathbf{Z}/(2^{32}-1)$ are shown to have many desirable properties, which makes them of potential interest for cryptographic applications. To further support the applications of this kind of sequences, in this paper, we consider the problem whether primitive sequences generated by two distinct primitive polynomials over $\mathbf{Z} /(2^{32}-1)$ are pairwise distinct modulo 2. A sufficient condition is given for ensuring that the answer to this problem is positive.

Appendices

Appendix 1: Proof of Proposition 1

In this section, we always assume that $p_{i}=2^{2^{i-1}}+1$ for $1\le i\le 5$. Then it is clear that $2^{32}-1=p_{1}p_{2}p_{3}p_{4}p_{5}$ is the canonical factorization of $2^{32}-1$. To prove Proposition 1, we first give some necessary lemmas.

Lemma 6

For $2\le j\le 5$ we have that

$$\begin{aligned} p_{1}\cdots p_{j-1} = p_{j}-2, \end{aligned}$$

(15)

$$\begin{aligned} p_{j}\cdot \frac{p_{1}\cdots p_{j-1}+1}{2} \equiv 1\left( \,\mathrm{mod}\,{p_{1}\cdots p_{j-1}}\right) . \end{aligned}$$

(16)

Proof

The result can be checked directly. $\square$

For any integer $m\ge 2$, let $v_{2}\left( m\right)$ denote the greatest nonnegative integer k such that $2^{k}$ divides m. Then it follows from [7, Lemma 10] that

$$\begin{aligned} v_{2}\left( \frac{p_{i}^{n}-1}{p_{i}-1}\right) =v_{2}\left( p_{i}^{n}-1\right) -2^{i-1}=\left\{ \begin{array}{l} v_{2}\left( n\right) +1,\quad \mathrm{if}\quad i=1\quad \mathrm{and}\quad n\quad \mathrm{is}\; \mathrm{even}; \\ v_{2}\left( n\right) , \quad \mathrm{otherwise}, \end{array} \right. \quad \quad \end{aligned}$$

(17)

and so

$$\begin{aligned} v_{2}\left( \mathrm{lcm}\left( \frac{p_{1}^{n}-1}{p_{1}-1},\ldots ,\frac{ p_{5}^{n}-1}{p_{5}-1}\right) \right) =\left\{ \begin{array}{l} v_{2}\left( n\right) +1,\quad \mathrm{if}\quad n\quad \mathrm{is}\; \mathrm{even}; \\ v_{2}\left( n\right) , \quad \mathrm{if}\quad n\quad \mathrm{is}\; \mathrm{odd}. \end{array} \right. \end{aligned}$$

(18)

Lemma 7

Let $\xi _{f}$ be the associated element of a primitive polynomial $f\left( x\right)$ of degree $n\ge 2$ over $\mathbf{Z} /(2^{32}-1)$. Then

$$\begin{aligned} ord\left( \xi _{f},p_{i}\right) =\left\{ \begin{array}{l} 2^{2^{i-1}-1},\quad \mathrm{if }2\le i\le 5\quad \mathrm{and}\quad n\quad \mathrm{is}\; \mathrm{even}; \\ 2^{2^{i-1}},\quad \mathrm{otherwise}, \end{array} \right. \end{aligned}$$

where $ord(\xi _{f},p_{i})$ is the multiplicative order of $\xi _{f}$ mod $p_{i}$.

Proof

It follows from 3 that

$$\begin{aligned} \xi _{f}\equiv \left( (-1)^{n}f(0)\right) ^{\frac{p_{i}-1}{p_{i}^{n}-1}\cdot \theta } \,\mathrm{mod}\,{p_{i}}\quad \mathrm{for}\;1\le i\le 5, \end{aligned}$$

where $[(-1)^{n}f(0)]_{\,\mathrm{mod}\,{p_{i}}}$ is a primitive element of $\mathbf{Z }/(p_{i})$ and $\theta =\mathrm{lcm}\left( \frac{p_{1}^{n}-1}{p_{1}-1},\ldots , \frac{p_{5}^{n}-1}{p_{5}-1}\right)$. Then we get that

$$\begin{aligned} ord\left( \xi _{f},p_{i}\right) =\frac{p_{i}-1}{\gcd \left( p_{i}-1,\,\frac{p_{i}-1}{ p_{i}^{n}-1}\cdot \theta \right) }=\frac{2^{2^{i-1}}}{\gcd \left( 2^{2^{i-1}},\,\frac{p_{i}-1}{p_{i}^{n}-1}\cdot \theta \right) }=2^{w_{i}}, \end{aligned}$$

(19)

where

$$\begin{aligned} w_{i}=2^{i-1}-\min \left\{ 2^{i-1},v_{2}(\theta )-v_{2}\left( \frac{ p_{i}^{n}-1}{p_{i}-1}\right) \right\} ,\quad 1\le i\le 5. \end{aligned}$$

(20)

Applying 17 and 18 to 20 we obtain that

$$\begin{aligned} w_{i}=\left\{ \begin{array}{l} 2^{i-1}-1,\quad \mathrm{if }\quad 2\le i\le 5\quad \mathrm{and}\quad n\quad \mathrm{is}\; \mathrm{even}; \\ 2^{i-1},\quad \mathrm{otherwise}. \end{array} \right. \end{aligned}$$

(21)

Then 19 and 21 yield

$$\begin{aligned} ord\left( \xi _{f},p_{i}\right) =\left\{ \begin{array}{l} 2^{2^{i-1}-1},\quad \mathrm{if}\quad 2\le i\le 5\quad \mathrm{and}\quad n\quad \mathrm{is}\; \mathrm{even}; \\ 2^{2^{i-1}},\quad \mathrm{otherwise}. \end{array} \right. \end{aligned}$$

$\square$

Lemma 8

Let $p\in \{5,17,257,65537\}$ and $\xi \in \mathbf{Z}/(p)$ with $ord(\xi ,p)=(p-1)/2$. If $\underline{a}\in G(x-\xi ,p)$ with $\underline{a}\ne \underline{0}$, then either 1 or 3 occurs in $\underline{a}$.

Proof

Set $\underline{b}=\left( [\xi ^{t}]_{\,\mathrm{mod}\,{p}}\right) _{t\ge 0}$. Then it is clear that $\underline{b}\in G(x-\xi ,p)$ with $per(\underline{b} )=(p-1)/2$. Since $L^{k}\underline{b}\in G(x-\xi ,p)$ for any integer $k\ge 0$, where $L^{k}\underline{b}=(b(k+t))_{t\ge 0}$ is the $k$-shift of $\underline{b}$, it implies that $\underline{b},L\underline{b},\ldots ,L^{ \frac{p-1}{2}-1}\underline{b}$ are $\frac{p-1}{2}$ distinct sequences in the set $G(x-\xi ,p)$.

Note that three is a primitive element of $\mathbf{Z}/(p)$, and so it does not occur in $L^{k}\underline{b}$ for any integer $0\le k\le \frac{p-1}{2}-1$ (since otherwise there exists an integer $t^{*}\ge 0$ such that $[ \xi ^{t^{*}}] _{\,\mathrm{mod}\,{p}}=3$, which implies that $\xi$ is a primitive element of $\mathbf{Z}/(p)$, a contradiction to the assumption that $ord(\xi ,p)=(p-1)/2$). Set $\underline{c}=\left( [3\cdot \xi ^{t}]_{ \,\mathrm{mod}\,{p}}\right) _{t\ge 0}$. Then $\underline{c}\in G(x-\xi ,p)$ with $per(\underline{c})=(p-1)/2$. Note that $c\left( 0\right) =3$, and so we have that

$$\begin{aligned} \underline{c}\ne L^{k}\underline{b}\quad \mathrm{for}\quad 0\le k\le \frac{p-1}{2}-1. \end{aligned}$$

Similar to the analyses above, $\underline{c},L\underline{c},\ldots ,L^{ \frac{p-1}{2}-1}\underline{c}$ are $\frac{p-1}{2}$ distinct sequences in the set $G(x-\xi ,p)\backslash \{\underline{b},L\underline{b},\ldots ,L^{\frac{p-1}{2}-1}\underline{b}\}$.

Since there are only p distinct sequences in the set $G(x-\xi ,p)$, we get that

$$\begin{aligned} G\left( x-\xi ,p\right) =\left\{ \underline{0},\underline{b},L\underline{b},\ldots ,L^{ \frac{p-1}{2}-1}\underline{b},\underline{c},L\underline{c},\ldots ,L^{\frac{ p-1}{2}-1}\underline{c}\right\} . \end{aligned}$$

Therefore either $\underline{a}=L^{k}\underline{b}$ or $\underline{a}=L^{k} \underline{c}$ for some $0\le k\le \frac{p-1}{2}-1$. Note that $b(0)=1$ and $c(0)=3$, and so either 1 or 3 occurs in $\underline{a}$. $\square$

Lemma 9

Let $\xi _{f}$ be the associated element of a primitive polynomial $f(x)$ of degree $n\ge 2$ over $\mathbf{Z}/(2^{32}-1)$ and $\underline{a}\in G(x-\xi _{f},2^{32}-1)$. If $\left[ \underline{a}\right] _{ \,\mathrm{mod}\,{p_{j}}}\ne \underline{0}$ for some $j\in \{1,2,3,4,5\}$, then

$$\begin{aligned} a(t+T/2)\equiv -a(t)\,\mathrm{mod}\,{p_{{j}}}\quad \mathrm{for}\quad t\ge 0, \end{aligned}$$

where $T=ord\left( \xi _{f},p_{j}\right)$. Moreover, if T is divisible by four, then

$$\begin{aligned} a(t+T/4)=\left[ \pm 2^{2^{j-2}}\cdot a(t)\right] _{\,\mathrm{mod}\,{p_{j}}}\quad \mathrm{for}\quad t\ge 0. \end{aligned}$$

Proof

Note that $ord(\xi _{f}^{T/2},p_{j})=2$ and $ord(\xi _{f}^{T/4},p_{j})=4$ if $4|T$, and so we get that

$$\begin{aligned} \xi _{f}^{T/2}\equiv -1\left( {\,\mathrm{mod}\,}{p_{j}}\right) \quad \mathrm{and}\quad \xi _{f}^{T/4}\equiv \pm 2^{2^{j-2}}\left( {\,\mathrm{mod}\,}{p_{j}}\right) \quad \mathrm{if}\; 4|T. \end{aligned}$$

(22)

Then the result immediately follows by applying 22 to $\underline{a}$. $\square$

Lemma 10

([8, Proposition 8]) Let p and q be two positive integers with $\gcd (p,q)=1$. Then

$$\begin{aligned} x=x_{1}+p\cdot \left[ p^{-1}\cdot \left( x_{2}-x_{1}\right) \right] _{\,\mathrm{mod}\,{q}}, \end{aligned}$$

is the unique integer in the half-open interval $[0,pq)$ such that

$$\begin{aligned}{}[ x]_{\,\mathrm{mod}\,{p}}=x_{1}\quad \mathrm{and}\quad [x]_{\,\mathrm{mod}\,{ q}}=x_{2}, \end{aligned}$$

where $p^{-1}$ is the multiplicative inverse of $p\,\mathrm{mod}\,{q}$.

With the preparations above, now we are ready to prove Proposition 1.

Proof of Proposition 1

Let $p_{j}$ be the largest prime divisor of $2^{32}-1$ such that $[ \underline{a}]_{\,\mathrm{mod}\,{p_{j}}}\ne \underline{0}$. Put $k=(2^{32}-1)/p_{1}\cdots p_{j}$. Then it is clear that

$$\begin{aligned} \underline{a}=k\cdot \underline{c}, \end{aligned}$$

where $\underline{c}\in G(x-[\xi _{f}]_{\,\mathrm{mod}\,{p_{1}\cdots p_{j}}},\,p_{1}\cdots p_{j})$ with $[\underline{c}]_{\,\mathrm{mod}\,{p_{j}}}\ne \underline{0}$. Note that $k$ is odd, and so to prove the proposition, it suffices to show that there is an integer $t^{*}\ge 0$ such that $c(t^{*})$ is an odd integer. Since the result is obviously true if $j=1$ (note that $[\xi _{f}]_{\,\mathrm{mod}\,{p_{1}}}$ is a primitive element of $\mathbf{Z}/(p_{1})$, and so $\underline{c}$ is a primitive sequence over $\mathbf{Z} /(p_{1})$), we only prove for the case $j>1$.

Put $q=p_{1}\cdots p_{j-1}$. Then by Lemma 10 together with 16, $\underline{c}$ can be rewritten as

$$\begin{aligned} \underline{c}=\underline{u}+p_{j}\cdot \left[ \frac{q+1}{2}\cdot \left( \underline{v}-\underline{u}\right) \right] _{\,\mathrm{mod}\,{q}}, \end{aligned}$$

(23)

where

$$\begin{aligned} \underline{u}=[\underline{c}]_{\,\mathrm{mod}\,{p_{j}}}\quad \mathrm{and}\quad \underline{v}=[\underline{c}]_{\,\mathrm{mod}\,{q}}. \end{aligned}$$

Denote $T_{u}=per(\underline{u})$ and $T_{v}=per(\underline{v})$. Note that $per(\underline{u})=ord(\xi _{f},p_{j})$ and $per(\underline{v})|ord(\xi _{f},p_{j-1})$, and so it follows from Lemma 7 that

$$\begin{aligned} T_{u}=\left\{ \begin{array}{l} 2^{2^{j-1}-1},\quad \mathrm{if}\quad n\quad \mathrm{is}\; \mathrm{even}; \\ 2^{2^{j-1}},\quad \mathrm{otherwise}, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} T_{v}\mid ord\left( \xi _{f},p_{j-1}\right) =\left\{ \begin{array}{l} 2,\quad \mathrm{if}\; j=2; \\ 2^{2^{j-2}-1},\quad \mathrm{if }\quad j\ge 3\quad \mathrm{and}\quad n\quad \mathrm{is}\; \mathrm{even}; \\ 2^{2^{j-2}},\quad \mathrm{otherwise}. \end{array} \right. \end{aligned}$$

It can be seen that

$$\begin{aligned} T_{v}\mid \frac{T_{u}}{2}\quad \mathrm{if}\quad n\quad \mathrm{is}\; \mathrm{odd}; \quad \mathrm{and}\quad T_{v}\mid \frac{ T_{u}}{4}\quad \mathrm{if}\; j\ge 3. \end{aligned}$$

(24)

Case 1: $n$ is odd.

Since, in this case, $T_{u}=2^{2^{j-1}}=p_{j}-1$, it implies that $\underline{u}$ is a primitive sequence of order 1 over $\mathbf{Z}/(p_{j})$, and so 1 occurs in $\underline{u}$. We assume without loss of generality that $u(0)=1$. Then by Lemma 9 we have $u(T_{u}/2)=p_{j}-1$. Therefore 23 yields that

$$\begin{aligned} c(0) =1+p_{j}\cdot \left[ \frac{q+1}{2}\cdot (v(0)-1)\right] _{\,\mathrm{mod}\,{q}}, \end{aligned}$$

(25)

$$\begin{aligned} c(T_{u}/2)=p_{j}-1+p_{j}\cdot \left[ \frac{q+1}{2}\cdot \left( v(T_{u}/2)-( p_{j}-1)\right) \right] _{\,\mathrm{mod}\,{q}}. \end{aligned}$$

(26)

Note that 24 and 15, respectively, imply that

$$\begin{aligned} v\left( T_{u}/2\right) =v(0)\quad \mathrm{and}\quad \left[ p_{j}-1\right] _{\,\mathrm{mod}\,{ q}}=1. \end{aligned}$$

(27)

Thus by applying 27 to 26 we get that

$$\begin{aligned} c\left( T_{u}/2\right) =p_{j}-1+p_{j}\cdot \left[ \frac{q+1}{2}\cdot (v(0)-1)\right] _{\,\mathrm{mod}\,{q}}. \end{aligned}$$

(28)

Then it can be seen from 25 and 28 that either $c(0)$ or $c(T_{u}/2)$ is an odd integer.

Case 2: $n$ is even.

Case 2.1: $n$ is even and $j\ge 3$.

We note that in this case $T_{v}\mid \frac{T_{u}}{4}$ and $T_{u}=2^{2^{j-1}-1}=(p_{j}-1)/2$. If 1 occurs in $\underline{u}$, then similar to Case 1, either $c(0)$ or $c(T_{u}/2)$ is an odd integer. If one does not occurs in $\underline{u}$, then it follows from Lemma 8 that three must occur in $\underline{u}$. Assume without loss of generality that $u(0)=3$. Similar to the analyses in Case 1, we can get that

$$\begin{aligned} u\left( T_{{u}}/2\right) =p_{j}-3=q-1\quad \mathrm{and}\quad v\left( T_{u}/2\right) =v(0), \end{aligned}$$

(29)

and so

$$\begin{aligned} c(0)&= 3+p_{j}\cdot \left[ \frac{q+1}{2}\cdot ( v(0)-3)\right] _{ \,\mathrm{mod}\,{q}} \\&= 3+p_{j}\cdot \left[ \frac{q+1}{2}\cdot v(0)-\frac{q+3}{2}\right] _{\,\mathrm{mod}\,{q}}, \\ c(T_{u}/2)&= q-1+p_{j}\cdot \left[ \frac{q+1}{2}\cdot \left( v(0)-( q-1)\right) \right] _{\,\mathrm{mod}\,{q}} \\&= q-1+p_{j}\cdot \left[ \frac{q+1}{2}\cdot v(0)-\frac{q-1}{2}\right] _{\,\mathrm{mod}\,{ q}}. \end{aligned}$$

If $[\frac{q+1}{2}\cdot v(0)]_{\,\mathrm{mod}\,{q}}\notin \{\frac{q-1}{2}, \frac{q+1}{2}\}$, then

$$\begin{aligned} \left[ \frac{q+1}{2}\cdot v(0)\right] _{\,\mathrm{mod}\,{q}}\ge \frac{q+3}{2}\text { or } \left[ \frac{q+1}{2}\cdot v(0)\right] _{\,\mathrm{mod}\,{q}}<\frac{q-1}{2}. \end{aligned}$$

In either case we have

$$\begin{aligned} \left[ \frac{q+1}{2}\cdot v(0)-\frac{q+3}{2}\right] _{\,\mathrm{mod}\,{q}}\equiv \left[ \frac{q+1}{2}\cdot v(0)-\frac{q-1}{2}\right] _{\,\mathrm{mod}\,{q}}\,\mathrm{mod}\,{2}, \end{aligned}$$

which implies that either $c(0)$ or $c(T_{u}/2)$ is an odd integer.

If $[\frac{q+1}{2}\cdot v(0)]_{\,\mathrm{mod}\,{q}}=\frac{q+1}{2}$, then it is clear that $v(0)=1$, and so

$$\begin{aligned} c(0)=3+p_{j}\cdot \left[ \frac{q+1}{2}-\frac{q+3}{2}\right] _{\,\mathrm{mod}\,{ q}}=3+p_{j}\cdot (q-1), \end{aligned}$$

which implies that $c(0)$ is an odd integer.

If $[\frac{q+1}{2}\cdot v(0)]_{\,\mathrm{mod}\,{q}}=\frac{q-1}{2}$, then it follows that $v(0)=q-1$, and so by 24 we get that

$$\begin{aligned} v\left( T_{u}/4\right) =v\left( 0\right) =q-1. \end{aligned}$$

Since $u(T_{u}/4)=[\pm 2^{2^{j-2}}\cdot u(0)]_{\,\mathrm{mod}\,{p_{j}}}=[\pm 2^{2^{j-2}}\cdot 3]_{\,\mathrm{mod}\,{p_{j}}}$ by Lemma 9, it follows 23 that

$$\begin{aligned} c\left( T_{u}/4\right) =\left[ \pm 2^{2^{j-2}}\cdot 3\right] _{\,\mathrm{mod}\,{p_{j}}}+p_{j}\cdot \left[ \frac{q+1}{2}\cdot \left( q-1-\left[ \pm 2^{2^{j-2}}\cdot 3\right] _{\,\mathrm{mod}\,{ p_{j}}}\right) \right] _{\,\mathrm{mod}\,{q}}. \end{aligned}$$

It can be verified that $c(T_{u}/4)$ is always an odd integer for any $j\in \left\{ 3,4,5\right\}$.

Case 2.2: $n$ is even and $j=2$.

In this case, it is clear that $p_{j}=5$, $q=3$ and $\underline{c}\in G(x-\left[ \xi _{f}\right] _{\,\mathrm{mod}\,{15}},15)$. Since by Lemma 7 we have that

$$\begin{aligned} ord\left( \xi _{f},3\right) =2\quad \mathrm{and}\quad ord\left( \xi _{f},5\right) =2, \end{aligned}$$

it implies that $\left[ \xi _{f}\right] _{\,\mathrm{mod}\,{3}}=2$ and $\left[ \xi _{f}\right] _{\,\mathrm{mod}\,{5}}=4$, and so $\left[ \xi _{f}\right] _{\,\mathrm{mod}\,{ 15}}=14$. It can be seen that $c(1)=[14\cdot c(0)]_{\,\mathrm{mod}\,{15}}=15-c(0)$ and $c(0)\ne 0$, thus either $c(0)$ or $c(1)$ is an odd integer. $\square$

Appendix 2

As a preparation, we first introduce a result of Bugeaud etc.

Lemma 11

([9, Theorem 1]) Let p and q be two distinct odd prime integers. Then for any given real number $\varepsilon >0$, there exists a positive integer $N_{\varepsilon }$ such that

$$\begin{aligned} \gcd \left( p^{n}-1,q^{n}-1\right) <2^{\varepsilon \cdot n}\quad \mathrm{for}\quad n>N_{\varepsilon }. \end{aligned}$$

Now we can make the result explicit in the following statement.

Lemma 12

There exists a positive integer $N^{*}$ such that inequalities 5 always hold for all $n>N^{*}$.

Proof

Given a real number $\varepsilon >0$, for any $1\le u<v\le 5$, it follows from Lemma 11 that there exists positive integers $N_{\varepsilon }^{(u,v)}$ such that

$$\begin{aligned} \gcd \left( p_{\sigma (u)}^{n}-1,\,p_{\sigma (v)}^{n}-1\right) <2^{\varepsilon \cdot n} \quad \mathrm{for}\quad n>N_{\varepsilon }^{(u,v)}. \end{aligned}$$

(30)

Set $N_{\varepsilon }=\max \{N_{\varepsilon }^{(u,v)}\mid 1\le u<v\le 5\}$. Then by 30 we have that

$$\begin{aligned} \gcd \left( p_{\sigma (i+1)}^{n}-1,\,p_{\sigma (j)}^{n}-1\right) <2^{\varepsilon \cdot n} \quad \mathrm{for}\quad n>N_{\varepsilon }\quad \mathrm{and}\quad 1\le j\le i\le 4 \end{aligned}$$

and so

$$\begin{aligned} \frac{p_{\sigma (i+1)}^{n}-1}{\gcd \left( p_{\sigma (i+1)}^{n}-1,\theta _{i}\right) }&\ge \frac{p_{\sigma (i+1)}^{n}-1}{\prod _{1\le j\le i}\gcd \left( p_{\sigma (i+1)}^{n}-1,\,\frac{p_{\sigma (j)}^{n}-1}{p_{\sigma (j)}-1}\right) }\nonumber \\&\ge \frac{p_{\sigma (i+1)}^{n}-1}{\prod _{1\le j\le i}\gcd \left( p_{\sigma (i+1)}^{n}-1,\,p_{\sigma (j)}^{n}-1\right) } \nonumber \\&> \frac{p_{\sigma (i+1)}^{n}-1}{2^{i\cdot \varepsilon \cdot n}} \nonumber \\&> \left( \frac{p_{\sigma (i+1)}}{2^{i\cdot \varepsilon }}\right) ^{n}-1. \end{aligned}$$

(31)

Choose $\varepsilon >0$ such that

$$\begin{aligned} \frac{p_{\sigma (i+1)}}{2^{i\cdot \varepsilon }}>p_{\sigma (i+1)}^{1/2}+1 \quad \mathrm{for}\quad 1\le i\le 4, \end{aligned}$$

(32)

then by 31 and 32 we obtain that

$$\begin{aligned} \frac{p_{\sigma (i+1)}^{n}-1}{\gcd \left( p_{\sigma (i+1)}^{n}-1,\theta _{i}\right) } >\left( p_{\sigma (i+1)}^{1/2}+1\right) ^{n}-1>p_{\sigma (i+1)}^{n/2} \end{aligned}$$

for $1\le i\le 4$ and $n>N_{\varepsilon }$. Thus $N^{*}=N_{\varepsilon }$ is the desired integer. This completes the proof. $\square$